Hypothesis Testing (One-Sample)Sample SizeSample MeanPopulation Std Dev (Given in the problem)Hypothesized ValueLevel of significanceStandard ErrorTest Statisticp-value (lower-tail)p-value (upper-tail)p-value (two-tail)Critical Value (lower-tail)Critical Value upper-tail)Critical Value (two-tail) Use absolute value if test stat is positiveN Dist (Summary Statistic)50297.6122950.051.6970562751.5320646930.9372467640.0627532360.125506471Numbers in this column are examples from the slides-1.6448536271.644853627-1.959963985N Dist (Raw Data)86.252.91547594760.051.0307764060.2425356250.5958174220.4041825780.808365156-1.6448536271.644853627-1.959963985NewBalance156784910 Exhibit 9-31n = 64x= 64.8o = 24Refer to Exhibit 9-31. The test statistic equalsO.396Ho: μ = 60H:μ*60O-1.201.21.6

Answered: Image /qna-images/answer/a5f606a8-cbd9-49cd-8f83-908420da049a.jpg

Answered: n = 64 X= 64.8 a=24 Ho: μ = 60 H₂:μ*60…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Z-value p-value Single Policyholders n₁ = 440 Number making claims = 88 a. Use a = 0.05. Test to determine whether the claim rates differ between single and married male policyholders. (to 2 decimals) (to 4 decimals) X Married Policyholders n2 = 840 Number making claims 126 We can conclude ✔ that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number. X * )
Suppose a new drug is intended to improve walking distance in patients with Parkinson Disease A researcher measures patients walking distance in feet before and after trying the drug for 9 weeks. What is the value of t obs
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Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Question: If a participant has the same corneal thickness in their normal eye as the eye with Glaucoma, what would be the value for difference: measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma.
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Question:If the t statistic falls below the effect size in which we reject the null, but the p value is greater than 0.05. Do i reject the null? As my scenario had no population mean because the differences were recorded, do i set a test value of 10 or keep it at 0?
Please answer
Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Question: Write a report summarizing your findings When writing the report consider that medical staff estimate that a difference of 4.5 microns or more could impact on their ability to interpret eye pressure correctly.
P-value?

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n = 64 X= 64.8 a=24 Ho: μ = 60 H₂:μ*60 Refer to Exhibit 9-31. The test statistic equals